CareerCup

public int countTrailingZerosOfFactorial(int n) {
    int count = 0;
    for (int powerOfFive = 5; n / powerOfFive > 0; powerOfFive *= 5) {
        count += n / powerOfFive;
    }
    return count;
}

24. Factorize multiples of 5
5 = 5x1
10 = 5x2
15 = 5x3
20 = 5x2x2
...
25 = 5x5 (one extra 5)
..
50 = 5x5x2 (one extra 5)
...
75 = 5x5x3 (one extra 5)
..
100 = 5x5x4 (one extra 5)

All of the 20 multiples of 5 up to 100 have one 5 each. The multiples 25,50,75 & 100 have one extra 5 each. Hence the total number of 5s = 24. Each of these twenty four 5s when multiplied by 2(which are at least 50 in number) yields a zero and hence the total number of 0s = 24.

The easiest way is to check the maximum power of 5 which can be there in factorial.

10=5x2
If we are calculating the factorial then n!=n x n-1 .... 1= A x 5^p x 2^q, hence it will depend on the value of p which will define the number of zeros.

Steps to find the max number of 5 is
If the number is say Z then max power of 5
p= z/(5) +z/(25) + z/(125)...z/(5^n) till z/(5^n) results a integer number.

int Zeros(int n)
{
int k=0,nb=0;
while(pow(5,k+1)<=n)
{
k=k+1;
nb=nb+n/pow(5,k);
}
return nb;
}

f(n) = n/5 + f(n/5);
base case: f(n)=0 iff n<5

public class CalculateNumberOfZeros {

	public static void main(String[] args) {
		
		findMaxPowerof5(100);
	}

	public static void findMaxPowerof5(int num) {
		LinkedList<Integer> link = new LinkedList<Integer>();
		int totPower = 0;
		for (int i = 1; i <= num; i++) {
			if (Math.pow(5, i) <= num) {
				totPower += num / Math.pow(5, i);
			}
		}

		System.out.println("Number of zero=" + totPower);

	}
}

In Excel, cell A1: =FACT(100) Result is 93326215443944200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

In cell B1: =LEN(A1)
Copy the significant digits out of A1 (933262154439442)
In cell A2: =A1/933262154439442
In cell B2: =LEN(A2)
Number of zeros = B1-B2=14

This question is worded incorrectly, there are 30 zeros contained in 100!

100! = 
9332621544394415268169923885626670049071596826438162146859 29638952175999932299156089414639761565182862536979208272237582511852109168 64000000000000000000000000

The question should be "How many consecutive zeros are there at the end of 100! (100 factorial)."

import java.util.*;
class FactMag
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
double no=0,fact=1;
System.out.println("enter a number");
no=sc.nextDouble();
double r=0,c=0;
for(double i=1;i<=no;i++)
{
fact=fact*i;
}
System.out.println(fact+" ");
while(fact>0)
{
r=fact%10;
if(r==0)
c++;
fact=fact/10;
}
System.out.println("no of zeros are "+c);
}
}

import java.util.*;
class FactMag
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
double no=0,fact=1;
System.out.println("enter a number");
no=sc.nextDouble();
double r=0,c=0;
for(double i=1;i<=no;i++)
{
fact=fact*i;
}
System.out.println(fact+" ");
while(fact>0)
{
r=fact%10;
if(r==0)
c++;
fact=fact/10;
}
System.out.println("no of zeros are "+c);
}
}

each divisible by 5 and divisible by 10 are 0 contributors. but 25,50 ,75, 100 are two zeros contributors there are 8 5s and 8 10s making it 16 + (4*2) so totally 24 zeroes

143

I think problem is about zeros contained in the 10 Factorial not just trailing 0's

x = 1
for i in range(1,101):
    x*=i
cnt = 0
for val in str(x):
    if int(val) == 0:
        cnt += 1
print cnt

the factorial program its giving answer output is 5=5*4*3*2*1=120
this problem solves it anyone.

def f(n):
	if n==1:
		return 0
	if n%10==0:
		c=0
		a=n
		while a%10==0:
			a=a/10
			c=c+1
		return c+f(n-1)
	if n%5==0:
		c=0
		a=n
		while a%5==0:
			a=a/5
			c=c+1
		return c+f(n-1)
	return f(n-1)

100! will have 23 zeroes